WARNING! This post contains a spoiler for Problem #8 listed at Project Euler. Do not read the rest of this post if you’re planning to attempt to solve the problem yourself.
Problem #8 was thankfully a great deal more interesting than the last problem. It goes as follows..
Find the greatest product of five consecutive digits in the 1000-digit number.
73167176531330624919225119674426574742355349194934 96983520312774506326239578318016984801869478851843 85861560789112949495459501737958331952853208805511 12540698747158523863050715693290963295227443043557 66896648950445244523161731856403098711121722383113 62229893423380308135336276614282806444486645238749 30358907296290491560440772390713810515859307960866 70172427121883998797908792274921901699720888093776 65727333001053367881220235421809751254540594752243 52584907711670556013604839586446706324415722155397 53697817977846174064955149290862569321978468622482 83972241375657056057490261407972968652414535100474 82166370484403199890008895243450658541227588666881 16427171479924442928230863465674813919123162824586 17866458359124566529476545682848912883142607690042 24219022671055626321111109370544217506941658960408 07198403850962455444362981230987879927244284909188 84580156166097919133875499200524063689912560717606 05886116467109405077541002256983155200055935729725 71636269561882670428252483600823257530420752963450
I’m keeping with my Haskell theme and abusing some of the nice feature of Functional Programming to nail this one.
Here’s my solution:
-- ord is required to convert chars to ints import Data.Char (ord) -- a string representation of the large number number :: String number = "73167176531330624919225119674426574742355349194934" ++ "96983520312774506326239578318016984801869478851843" ++ "85861560789112949495459501737958331952853208805511" ++ "12540698747158523863050715693290963295227443043557" ++ "66896648950445244523161731856403098711121722383113" ++ "62229893423380308135336276614282806444486645238749" ++ "30358907296290491560440772390713810515859307960866" ++ "70172427121883998797908792274921901699720888093776" ++ "65727333001053367881220235421809751254540594752243" ++ "52584907711670556013604839586446706324415722155397" ++ "53697817977846174064955149290862569321978468622482" ++ "83972241375657056057490261407972968652414535100474" ++ "82166370484403199890008895243450658541227588666881" ++ "16427171479924442928230863465674813919123162824586" ++ "17866458359124566529476545682848912883142607690042" ++ "24219022671055626321111109370544217506941658960408" ++ "07198403850962455444362981230987879927244284909188" ++ "84580156166097919133875499200524063689912560717606" ++ "05886116467109405077541002256983155200055935729725" ++ "71636269561882670428252483600823257530420752963450" -- converts the number to a list of Int values numberList :: [Int] numberList = map (\n -> ord n - ord '0') number -- breaks up the number list into chunks of 5 digits chunks :: [[Int]] chunks = c' numberList 5 995 where c' _ _ 0 = [] c' l n c = (take n l) : c' (tail l) n (c - 1) -- main function iterates over the list of chunks, -- calculates the product of each of the chunks and -- then determines the biggest result main :: IO () main = print $ foldr1 max (map product chunks)
The comments are pretty descriptive of what goes on. It turns out that the solution isn’t that shabby in the performance stakes either:
Thu Jul 24 20:46 2008 Time and Allocation Profiling Report (Final)
main +RTS -p -RTS
total time = 0.00 secs (0 ticks @ 20 ms)
total alloc = 571,224 bytes (excludes profiling overheads)
COST CENTRE MODULE %time %alloc
CAF Main 0.0 99.8
individual inherited
COST CENTRE MODULE no. entries %time %alloc %time %alloc
MAIN MAIN 1 0 0.0 0.0 0.0 100.0
CAF Main 152 15 0.0 99.8 0.0 99.8
CAF GHC.Handle 88 4 0.0 0.2 0.0 0.2I still love how concise and easy the solutions to problems like this are in Haskell.
July 24, 2008
Hey there, this will save you the import:
– converts the number to a list of Int values
numberList :: [Int]
numberList = map (read . return)
July 24, 2008
Thanks Arnar! I have no idea why I didn’t think of that
Great advice. Cheers!